Divide the following rational expressions and simplify the result. $\dfrac{3y-5x}{4y^2-40xy}\div\dfrac{5xy}{y^2-20xy+100x^2}=$
Solution: Let's first factor the numerators and denominators of each expression separately. [Why are we doing this?] The numerator, $3y-5x$, of the dividend cannot be factored further. The denominator, $4y^2-40xy$, of the dividend can be factored as $4y(y-10x)$ by factoring out $4y$. The numerator, $5xy$, of the divisor cannot be factored further. The denominator, $y^2-20xy+100x^2$, of the divisor can be factored as $(y-10x)(y-10x)$ using the perfect square pattern. Now the division looks as follows: $\dfrac{3y-5x}{4y(y-10x)}\div\dfrac{5xy}{(y-10x)(y-10x)}$ To divide two rational expressions, we flip the divisor, multiply across, then simplify: [What's that?] $\begin{aligned} &\phantom{=} \dfrac{3y-5x}{4y(y-10x)}\div\dfrac{5xy}{(y-10x)(y-10x)} \\\\\\ &= \dfrac{3y-5x}{4y(y-10x)}\cdot \dfrac{(y-10x)(y-10x)}{5xy} &\text{Flip the divisor.}\\\\\\\\ &= \dfrac{(3y-5x) \cdot (y-10x)(y-10x)}{4y(y-10x) \cdot 5xy} &\text{Multiply across.}\\\\\\\\ &= \dfrac{(3y-5x) {\cancel{(y-10x)}}(y-10x)}{4y{\cancel{(y-10x)}} 5xy} &\text{Cancel out common factors.}\\\\\\\\ &= \dfrac{(3y-5x)(y-10x)}{20xy^2} \end{aligned}$ Therefore, the simplified form of the quotient is $\dfrac{(3y-5x)(y-10x)}{20xy^2}$, which is equivalent to $\dfrac{3y^2-35xy+50x^2}{20xy^2}$.